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16-1
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 14
Kinetics: Rates and Mechanisms of Chemical Reactions
16-2
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Figure 16.1 Reaction rate: the central focus of chemical kinetics
16-3
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Figure 16.2 The wide range of reaction rates.
16-4
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Factors That Influence Reaction Rate
Under a specific set of conditions, every reaction has its owncharacteristic rate, which depends upon the chemical nature ofthe reactants.
Five factors can be controlled during the reaction:
1. Concentration - molecules must collide to react;
2. Physical state - molecules must mix to collide;
3. Surface area – if reactants are in different phases, reaction occurs
where these phases meet
4. Temperature - molecules must collide with enough energy to react;
5. The use of a catalyst.
16-5
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Figure 16.3 The effect of surface area on reaction rate.
16-6
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Figure 16.4 Collision energy and reaction rate.
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Expressing the Reaction Rate
reaction rate - changes in the concentrations of reactants orproducts per unit time
reactant concentrations decrease while product concentrationsincrease
rate of reaction = -
for A B
change in concentration of A
change in time= -
conc A2-conc A1
t2-t1
(conc A)-
t
16-8
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Table 16.1 Concentration of O3 at Various Time in its
Reaction with C2H4 at 303K
C2H4(g) + O3(g) C2H4O(g) + O2(g)
Time (s) Concentration of O3 (mol/L)
0.0
20.0
30.0
40.0
50.0
60.0
10.0
3.20x10-5
2.42x10-5
1.95x10-5
1.63x10-5
1.40x10-5
1.23x10-5
1.10x10-5
(conc A)-
t
16-9
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Figure 16.5
The concentrations of O3 vs. time during its reaction with C2H4
C2H4(g) + O3(g) C2H4O(g) + O2(g)
- [C2H4]
t
rate =
- [O3]
t
=
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Figure 16.6 Plots of [C2H4] and [O2] vs. time.
Tools of the Laboratory
16-11
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Table 16.2 Initial Rates for a Series of Experiments in the Reaction Between O2 and NO
Experiment
Initial Reactant Concentrations (mol/L)
Initial Rate (mol/L*s)
1
2
3
4
5
O2 NO
1.10x10-2 1.30x10-2 3.21x10-3
1.10x10-2 3.90x10-2 28.8x10-3
2.20x10-2
1.10x10-2
3.30x10-2
1.30x10-2
2.60x10-2
1.30x10-2
6.40x10-3
12.8x10-3
9.60x10-3
2NO(g) + O2(g) 2NO2(g)
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Determining Reaction Orders
Using initial rates -
Run a series of experiments, each of which starts with a different set of reactant concentrations, and from each obtain an initial rate.
See Table 16.2 for data on the reaction
O2(g) + 2NO(g) 2NO2(g) rate = k [O2]m[NO]n
Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant.
k [O2]2m[NO]2
n
k [O2]1m[NO]1
n=
rate2
rate1 =
[O2]2m
[O2]1m
=
6.40x10-3mol/L*s
3.21x10-3mol/L*s
[O2]2
[O2]1
m
=1.10x10-2mol/L
2.20x10-2mol/L m; 2 = 2m m = 1
Do a similar calculation for the other reactant(s).
16-13
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Sample Problem 16.3
PLAN:
SOLUTION:
Determining Reaction Order from Initial Rate Data
PROBLEM: Many gaseous reactions occur in a car engine and exhaust system. One of these is
NO2(g) + CO(g) NO(g) + CO2(g) rate = k[NO2]m[CO]n
Use the following data to determine the individual and overall reaction orders.
Experiment Initial Rate(mol/L*s) Initial [NO2] (mol/L) Initial [CO] (mol/L)
1
2
3
0.0050
0.080
0.0050
0.10
0.10
0.40
0.10
0.10
0.20
Solve for each reactant using the general rate law using the method described previously.
rate = k [NO2]m[CO]n
First, choose two experiments in which [CO] remains constant and the [NO2] varies.
16-14
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Sample Problem 16.3 Determining Reaction Order from Initial Rate Data
continued
0.080
0.0050
rate 2
rate 1
[NO2] 2
[NO2] 1
m=
k [NO2]m2[CO]n
2
k [NO2]m1 [CO]n
1
=
0.40
0.10=
m; 16 = 4m and m = 2
k [NO2]m3[CO]n
3
k [NO2]m1 [CO]n
1
[CO] 3
[CO] 1
n=
rate 3
rate 1=
0.0050
0.0050=
0.20
0.10
n; 1 = 2n and n = 0
The reaction is 2nd order in NO2.
The reaction is zero order in CO.
rate = k [NO2]2[CO]0 = k [NO2]2
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Table 16.3 Units of the Rate Constant k for Several Overall Reaction Orders
Overall Reaction Order Units of k (t in seconds)
0 mol/L*s (or mol L-1 s-1)
1 1/s (or s-1)
2 L/mol*s (or L mol -1 s-1)
3 L2 / mol2 *s (or L2 mol-2 s-1)
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Figure 16.14
The effect of temperature on the distribution of collision energies
16-17
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Table 16.6 Rate Laws for General Elementary Steps
Elementary Step Molecularity Rate Law
A product
2A product
A + B product
2A + B product
Unimolecular
Bimolecular
Bimolecular
Termolecular
Rate = k [A]
Rate = k [A]2
Rate = k [A][B]
Rate = k [A]2[B]
REACTION MECHANISMS
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Sample Problem 16.8
PLAN:
SOLUTION:
Determining Molecularity and Rate Laws for Elementary Steps
PROBLEM: The following two reactions are proposed as elementary steps in the mechanism of an overall reaction:
(1) NO2Cl(g) NO2(g) + Cl (g)
(2) NO2Cl(g) + Cl (g) NO2(g) + Cl2(g)
(a) Write the overall balanced equation.
(b) Determine the overall rxn order for each step.
(a) The overall equation is the sum of the steps.(b) The molecularity is the sum of the reactant particles in the step.
2NO2Cl(g) 2NO2(g) + Cl2(g)
(c) Write the rate law for each step.
rate2 = k2 [NO2Cl][Cl]
(1) NO2Cl(g) NO2(g) + Cl (g)
(2) NO2Cl(g) + Cl (g) NO2(g) + Cl2(g)
(a)
Step(1) is unimolecular.Step(2) is bimolecular.
(b)
rate1 = k1 [NO2Cl](c)
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The Rate-Determining Step of a Reaction Mechanism
The overall rate of a reaction is related to the rate of the slowest, or rate-determining step.
Correlating the Mechanism with the Rate Law
The elementary steps must add up to the overall equation.
The elementary steps must be physically reasonable.
The mechanism must correlated with the rate law.